In Section 9.5 of Fangyang Zheng's Complex Differential Geometry Book, he proves the following:
Lemma 9.25. Let $(M^2,h)$ be a Kähler surface and $p \in M$. Suppose $M$ has negative holomorphic sectional curvature at $p$. Then there exists a basis $\{ e_1, e_2 \}$ of $T_p M$ such that $R_{1 \overline{1} 1 \overline{2}} = R_{2 \overline{2} 1 \overline{2}}=0$.
The argument is quite neat: Let $f : (\mathbb{P}^1 \times \mathbb{P}^1) - \Delta \to \mathbb{R}$ be the map $f([v],[w]) := R_{v \overline{v} w \overline{w}}/| v \wedge w |^2$, where $\mathbb{P}^1 = \mathbb{P}(T_pM)$. Since the holomorphic sectional curvature is negative at $p$, $f \to - \infty$ near the diagonal $\Delta$. Since $\mathbb{P}^1 \times \mathbb{P}^1$ is compact, with the maximum occurring away from $\Delta$, we choose $e_1, e_2 \in \mathbb{P}^1$ such that $\max(f) = R_{1 \overline{1} 2 \overline{2}}$. Zheng concludes with the assertion that the vanishing of the gradient implies the identity we want.
This last line is where I am confused. Since the maximum occurs at $v = e_1$ and $w = e_2$, if we compute the gradient, we have $\nabla_{e_1} R_{1 \overline{1} 2 \overline{2}} = 0$ and $\nabla_{e_2} R_{1 \overline{1} 2 \overline{2}}=0$. The second Bianchi identity only tells us that $0= \nabla_{e_1} R_{1 \overline{1} 2 \overline{2}} = \nabla_{e_2} R_{1 \overline{1} 1 \overline{2}}=0$, however, not that it vanishes.
I'm likely missing something completely trivial, so I apologise in advance.
